Actions triggered by field change in Django

Question

How do I have actions occur when a field gets changed in one of my models? In this particular case, I have this model:

class Game(models.Model):     STATE_CHOICES = (         ('S', 'Setup'),         ('A', 'Active'),         ('P', 'Paused'),         ('F', 'Finished')         )     name = models.CharField(max_length=100)     owner = models.ForeignKey(User)     created = models.DateTimeField(auto_now_add=True)     started = models.DateTimeField(null=True)     state = models.CharField(max_length=1, choices=STATE_CHOICES, default='S') 

and I would like to have Units created, and the 'started' field populated with the current datetime (among other things), when the state goes from Setup to Active.

I suspect that a model instance method is needed, but the docs don't seem to have much to say about using them in this manner.

Update: I've added the following to my Game class:

    def __init__(self, *args, **kwargs):         super(Game, self).__init__(*args, **kwargs)         self.old_state = self.state      def save(self, force_insert=False, force_update=False):         if self.old_state == 'S' and self.state == 'A':             self.started = datetime.datetime.now()         super(Game, self).save(force_insert, force_update)         self.old_state = self.state 

Answer 1

It has been answered, but here's an example of using signals, post_init and post_save.

from django.db.models.signals import post_save, post_init  class MyModel(models.Model):     state = models.IntegerField()     previous_state = None      @staticmethod     def post_save(sender, instance, created, **kwargs):         if instance.previous_state != instance.state or created:             do_something_with_state_change()      @staticmethod     def remember_state(sender, instance, **kwargs):         instance.previous_state = instance.state  post_save.connect(MyModel.post_save, sender=MyModel) post_init.connect(MyModel.remember_state, sender=MyModel)